Can I get a hand with this question about trig? I've really got no idea how to start.
Thanks guys.
Hey Yertle the Turtle,
Any time you have the sum of two inverse trig functions, it's a good sign that angle addition formulae may be involved. For part a:
\begin{align}
\tan^{-1}(\frac{1}{m}) + \tan^{-1}(\frac{1}{n}) &= \frac{\pi}{4}\\
\implies \tan(\tan^{-1}(\frac{1}{m}) + \tan^{-1}(\frac{1}{n})) &= \tan(\frac{\pi}{4})\\
\implies \frac{\tan(\tan^{-1}(\frac{1}{m})) + \tan(\tan^{-1}(\frac{1}{n}))}{1-\tan(\tan^{-1}(\frac{1}{m})) \tan(\tan^{-1}(\frac{1}{n}))} &= 1 \text{ (using angle addition on LHS)}\\
\implies \frac{\frac{1}{m} + \frac{1}{n}}{1-\frac{1}{mn}} &= 1\\
\implies \frac{\frac{n+m}{mn}}{\frac{mn-1}{mn}} &= 1\\
\implies n+m &= mn-1\\
\implies mn-m-n &= 1\\
\implies mn-m-n+1 &= 2\\
\therefore (m-1)(n-1) &= 2 \text{, as required}\\
\end{align}
Part b can be solved by finding n in terms of k, and then substituting into the equation given in the question.